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2.5 Just in case ...

... someone has forgotten how to find the extremum of a many-variable function under additional constraints. To remind you: the method of undetermined Lagrange multipliers waits to be applied here.

Let $f(x,y)=x^{2}+y^{2}$ be the given function. Of course, this is the equation of a parabolid with its tip - or minimum - at the origin. However, let $g(x,y) \equiv x-y-1=0$ be the constraint equation, meaning that we don't search for the global minimum but for the minimum along the line $y=x-1$. There are two ways to go about it. The simple but inconvenient way is to substitute $y=x-1$ in $f(x,y)$, thus rendering $f$ a function of $x$ only. Equating the derivative $df/dx$ to zero we find the locus of the conditional minimum, $x=1/2$ and $y=-1/2$. The process of substitution is, in general, tedious.

A more elegant method is this: defining a (undetermined) Lagrange multiplier $\alpha$, find the minimum of the function $f(x,y)-\alpha   g(x,y)$ according to

    $\displaystyle \frac{\partial f}{\partial x} - \alpha   \frac{\partial g}{\partial x} =0$ (2.53)
    $\displaystyle \frac{\partial f}{\partial y} - \alpha   \frac{\partial g}{\partial y} =0$ (2.54)

Eliminating $\alpha$ we find the solution without substituting anything. In our case
    $\displaystyle 2x-\alpha =0$ (2.55)
    $\displaystyle 2y+\alpha =0$ (2.56)

$\Longrightarrow$$x+y=0$, and from $g(x,y)=0$: $(x,y)=(1/2,-1/2)$.
next up previous
Next: 2.6 Problems for Chapter Up: 2. Elements of Kinetic Previous: 2.4 Transport processes
Franz Vesely
2005-01-25